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vodka [1.7K]
4 years ago
11

How much work would be needed to raise the payload from the surface of the moon (i.e., x = r) to the "end of the universe"?

Physics
2 answers:
meriva4 years ago
6 0
Moon was rotation Earth or sun
it's own orbit . universe is not end
marysya [2.9K]4 years ago
3 0

Answer:

W(3R) - W(2R) = -PR² (1/(3R) - 1/(2R)) = PR/6

Explanation:

"Assume the weight move up at constant speed. With no net acceleration, the force applied is -(weight). Since the weight at height R is -P(R/x)² (minusbecause it's directed downward) the applied lifting force is P(R/x)², and the work done moving from x to x+dx is dW = P(R/x)² dx. Intetgrate this:"

W(x) = ?PR²/x² dx = -PR²/x + C

 The work done moving from x=R to x=2R is:

W(2R) - W(R) = -PR²(1/(2R) - 1/R) = PR/2

(b) The work done moving from 2R to 3R is:

W(3R) - W(2R) = -PR² (1/(3R) - 1/(2R)) = PR/6

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4 years ago
The magnitudes of two forces are measured to be 120 ± 5 N and 60 ± 3 N. Find the sum and diff erence of the two magnitudes, givi
Pani-rosa [81]

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62N

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4 0
3 years ago
Sound waves are a- Transverse b- Longitudinal c- Transverse and longitudinal
Katen [24]

Answer:

Sound waves are longitudinal in nature.

Explanation:

There are many types of waves like transverse, longitudinal, electromagnetic wave etc.

Sound waves are longitudinal in nature. In longitudinal type of wave, the medium particles moves parallel to the propagation of the wave. This type of waves move in the form of compression and rarefaction.

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So, the correct option is (b) "longitudinal wave".

3 0
3 years ago
Read 2 more answers
Use this formula to solve this problem: Power (P) = Work (W) ÷ time (t) Peter's body supplies a force of 500 N to run up a 5 met
sesenic [268]
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4 0
3 years ago
A spherical snowball is melting in such a way that its radius is decreasing at a rate of 0.1 cm/min. At what rate is the volume
Ne4ueva [31]

Answer: 80.384 cubic cm /min

Explanation:

Let V denote the volume and r denotes the radius of the spherical snowball .

Given : \dfrac{dr}{dt}=-0.1\text{cm/min}

We know that the volume of a sphere is given by :-

V=\dfrac{4}{3}\pi r^3

Differentiating on the both sides w.r.t. t (time) ,w e get

\dfrac{dV}{dt}=\dfrac{4}{3}\pi(3r^2)\dfrac{dr}{dt}\\\\\Rightarrow\ \dfrac{dV}{dt}=4\pi r^2 (-0.1)=-0.4\pi r^2

When r= 8 cm

\dfrac{dV}{dt}=-0.4(3.14)(8)^2=-80.384

Hence, the volume of the snowball decreasing at the rate of 80.384 cubic cm /min.

6 0
3 years ago
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